Quizzes 16,17,19,24,25, and 26
How can we ascertain the number of polygenes involved in the inheritance of a quantitative trait?
(2n+1)(2n+1), where (2n+1) is the number of possible phenotypes and n equals the number of additive loci 1/4n1/4n, where n=the number of polygenes if the raio of F2 individuals resembling either of the two extreme P1 phenotypes can be determined
From the same data set, report the correlation coefficient.
-0.640
The weights and lengths of a sample of avocados are tabulated below. Mean weight (g) Length (cm) 169 8.85 152 9.41 177 7.63 102 9.79 154 6.78 210 5.44 195 8.26 189 8.57 Report the covariance between avocado weight and length.
-30.7
Considering the same population of cats as in Part A, what is the expected frequency of each genotype (TLTL, TLTS, TSTS ) based on the equation for Hardy-Weinberg equilibrium? Keep in mind that you just learned in Part A that: The allele frequency of TL is 0.4. The allele frequency of TS is 0.6. The equation for Hardy-Weinberg equilibrium states that at a locus with two alleles, as in this cat population, the three genotypes will occur in specific proportions: p2+2pq+q2=1p2+2pq+q2=1 For help applying the Hardy-Weinberg equation to this cat population, see Hints 1 and 2. Enter the values for the expected frequency of each genotype, TLTL, TLTS, TSTS, respectively, separated by commas. Enter your answers numerically (as decimals), not as percentages.
0.16,0.48,0.36
Use the information from Parts B and C to solve this problem.If a rabbit of genotype AaBb is bred to a rabbit of genotype AaBb, what is the probability of producing an offspring with an ear length of 13.5 cm?
0.250
Consider a population in which the frequency of allele AA is p=0.7p=0.7 and the frequency of allele aa is q=0.3q=0.3, and where the alleles are codominant. What will be the allele frequencies after one generation if the following occurs? wAA=1wAA=1 , wAa=0.9wAa=0.9 , and waa=0.8waa=0.8 . Enter your answers in the following order: qg+lqg+l , pg+lpg+l .
0.278, 0.722
wAA=1wAA=1 , wAa=0.95wAa=0.95 , and waa=0.9waa=0.9 . Enter your answers in the following order: qg+lqg+l , pg+lpg+l .
0.289,0.711
wAA=1wAA=1 , wAa=0.99wAa=0.99 , and waa=0.98waa=0.98 . Enter your answers in the following order: qg+lqg+l , pg+lpg+l .
0.298, 0.702
wAA=0.8wAA=0.8 , wAa=1wAa=1 , and waa=0.8waa=0.8 . Enter your answers in the following order: qg+lqg+l , pg+lpg+l .
0.319, 0.681
Calculate the realized heritability for 1997 year.
0.531
The mean and variance of plant height of two highly inbred strains (P1 and P2) and their progeny (F1 and F2) are shown here. StrainMean (cm)VarianceP138.04.3P260.02.0F149.04.8F251.19.5 Calculate the broad-sense heritability (H2 ) of plant height in this species.
0.58
Calculate the broad sense heritability using the previous two calculations, where: H2 = COVx'x"/Vx
0.837
Using the data from these two arrays, calculate the correlation coefficient. Remember this is a sample.
0.990
The operon model describes how bacteria control the production of groups of enzymes. In this model, synthesis of the messenger RNA coding for these enzymes is switched on or off by regulatory proteins. Can you match terms related to operons to their definitions?
1. A(n) operonis a stretch of DNA consisting of an operator, a promoter, and genes for a related set of proteins, usually making up an entire metabolic pathway. 2. Thegenes of an operonis/are arranged sequentially after the promoter. 3. A(n) promoteris a specific nucleotide sequence in DNA that binds RNA polymerase, positioning it to start transcribing RNA at the appropriate place. 4. A(n) regulatory genecodes for a protein, such as a repressor, that controls the transcription of another gene or group of genes. 5. Regulatory proteins bind to the operatorto control expression of the operon. 6. A(n) repressoris a protein that inhibits gene transcription. In prokaryotes, this protein binds to the DNA in or near the promoter. 7. A(n) induceris a specific small molecule that binds to a bacterial regulatory protein and changes its shape so that it cannot bind to an operator, thus switching an operon on.
Erma and Harvey were a compatible barnyard pair, but a curious sight. Harvey's tail was only 6 cmcm long, while Erma's was 30 cmcm. Their F1F1 piglet offspring all grew tails that were 18 cmcm. When inbred, an F2F2 generation resulted in many piglets (Erma and Harvey's grandpigs), whose tails ranged in 4-cmcm intervals from 6 to 30 cmcm (6, 10, 14, 18, 22, 26, and 30). Most had 18-cmcm tails, while 1/641/64 had 6-cmcm tails and 1/641/64 had 30-cmcm tails. Explain how these tail lengths were inherited by describing the mode of inheritance, indicating how many gene pairs were at work, and designating the genotypes of Harvey, Erma, and their 18-cmcm-tail offspring.
1. It is likely that there are three gene pairs in this cross. There are seven categories of phenotypes. The genotypes of the parents would be combinations of alleles that would produce a 6 cm (aabbcc) tail and a 30 cm (AABBCC) tail, whereas the 18 cm offspring would have a genotype of AaBbCc
If the 18−cmF118−cmF1 pig with genotype AABbcc is mated with one of the 6−cmF26−cmF2 pigs, what phenotypic ratio would be predicted if many offspring resulted?
14 cm/ 10 cm = 1 :1
Consider a true-breeding plant, AABBCCAABBCC, crossed with another true-breeding plant, aabbccaabbcc, whose resulting offspring are AaBbCcAaBbCc. If you cross the F1F1 generation, and independent assortment occurs, the expected number of offspring in each phenotypic class is given by the expression N!/[M!(N−M)!]N!/[M!(N−M)!] where NN is the total number of alleles (six in this example) and MM is the number of uppercase alleles. In the F1F1 cross of AaBbCb×AaBbCcAaBbCb×AaBbCc, how many offspring would be expected to contain 2 uppercase alleles?
15 out of 64 offspring
Indicate the possible genotypes that could account for F2F2 plants of various heights.
18cm : AaBbccdd aaBbCcdd AAbbccdd 33 cm: AABBCCDd, AABbCCDD AABBCcDD None: aabbCCDD, AABBCCDD, AaBbCcDd
Floral traits in plants often play key roles in diversification, in that slight modifications of those traits, if genetically determined, may quickly lead to reproductive restrictions and evolution. Insight into genetic involvement in flower formation is often acquired through selection experiments that expose realized heritability. Lendvai and Levin (2003) conducted a series of artificial selection experiments on flower size (diameter) in Phlox drummondii. Data from their selection experiments are presented in the following table in modified form and content. YearTreatmentMean (mm)1997Control30.04Selected parents34.13Offspring32.211998Control28.11Selected parents31.98Offspring31.901999Control29.68Selected parents31.81Offspring33.74 Considering that differences in control values represent year-to-year differences in greenhouse conditions, calculate (in mmmm) the average response to selection over the three-year period.
3.34 mm
Assume that a recessive autosomal disorder occurs in 1 of 25,000 individuals (0.00004) in the general population and that in this population about 4 percent (0.04) of the individuals are carriers for the disorder. Estimate the probability of this disorder occurring in the offspring of a marriage between first cousins.
5.00×10−3
Applying Hardy-Weinberg equilibrium, how many genotypes are predicted for a gene that has three alleles?
6
In terms of evolutionary potential, is a population with high heritability likely to be favored compared to one with a low realized heritability?
A population with a high heritability is likely to be favored.
Researchers collecting data on an island population of rabbits have found that the mutation rate of the recessive allele, d, is 8 x10-5. Also, the relative fitness of the phenotype encoded by dd, has been found to be 0.80.If the researchers let this rabbit population come to equilibrium, determine the allele frequency of the recessive allele.
Allele frequency of d at equilibrium = 2.00×10−2
In a different population, suppose that 38.40% of people are blood type A, and 46.24% are blood type O. Assuming this population is in Hardy-Weinberg equilibrium, what percentage of the population is blood type B? What percentage is blood type AB? Enter your answers to two decimal places.
B = 11.52% AB = 3.84%
A hypothetical population of 300 wolves has two alleles, FB and FW, for a locus that codes for fur color. The table below describes the phenotype of a wolf with each possible genotype, as well as the number of individuals in the population with each genotype. Which statements accurately describe the population of wolves? FBFBblack 40FBFWgray 40FWFWwhite 220
Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFBgenotype is 12. Based on the equation for Hardy-Weinberg equilibrium, the expected number of wolves with the FBFWgenotype is 96. The population may be evolving because the actual number of individuals with each genotype differs from the expected number of individuals with each genotype. The population is not at Hardy-Weinberg equilibrium.
In a sample population of chickens, genotype BB has a relative fitness of 1.0. Genotype Bb has a selection coefficient of 0.3, and genotype bb has a selection coefficient of 0.4.What is the relative fitness of genotype Bb? What is the relative fitness of genotype bb?
Bb, bb = 0.700,0.600
This table shows the frequency of given genotypes before and after selection. Genotype:CCCcccFrequency0.200.500.30Survivors after selection (total = 0.80)00.500.30 Identify the genotype(s) with a relative fitness of 1.0. Select all that apply.
Cc, cc
A developmental disorder in humans called spina bifida is a neural tube defect linked to a maternal diet low in folate during pregnancy. What does this suggest about the cause of spina bifida?
Folic acid is used by the body in several important processes including serving as a methylation precursor. Low levels of this substance are correlated with occurrence of the disorder, which suggests that spina bifida is a(n) epigenetic disorder with a(n) environmental root.
The Hardy-Weinberg principle states that, if a population is not evolving, then the frequencies of alleles and genotypes in that population will remain constant from one generation to the next. Further, this principle allows us to predict what the genotype frequencies will be in a non-evolving population. We can conclude that a population may be evolving if its genotype frequencies differ from those predicted by the Hardy-Weinberg principle. A hypothetical population of 200 cats has two alleles, TL and TS, for a locus that codes for tail length. The table below describes the phenotypes of cats with each possible genotype, as well as the number of individuals in the population with each genotype. Which statements about the population are true? TLTLlong 60TLTSmedium 40TSTSshort 100
Heterozygotes make up 20% of the population. Homozygotes make up 80% of the population. In the entire cat population, 60% of the alleles are TS. In the entire cat population, the frequency of the TL allele is 0.4. Assuming random mating, each gamete has a 40% chance of having a TL allele and a 60% chance of having a TS allele.
Compare this probability to the population at large.
Inbreeding increases the probability of this disorder.
A recent study examining the mutation rates of 5669 mammalian genes (17,208 sequences) indicates that, contrary to popular belief, mutation rates among lineages with vastly different generation lengths and physiological attributes are remarkably constant (Kumar, S., and Subramanian, S. 2002. Proc. Natl. Acad. Sci. [USA] 99: 803-808). The average rate is estimated at 12.2×10−912.2×10−9 per bp per year. What is the significance of this finding in terms of mammalian evolution?
It provides for increased understanding of the mutational processes. It provides more credible estimates of divergence times of species. It allows for broader interpretations of sequence comparisons.
Amino acids are classified as positively charged, negatively charged, or electrically neutral. Which category includes lysine?
Lysine is a positively charged amino acid.
DNA methylation is commonly associated with a reduction of transcription. The following data come from a study of the impact of the location and extent of DNADNA methylation on gene activity in eukaryotic cells. A bacterial gene, luciferase, was inserted into plasmids next to eukaryotic promoter fragments. CpG sequences, either within the promoter and coding sequence (transcription unit) or outside of the transcription unit, were methylated to various degrees, in vitro. The chimeric plasmids were then introduced into cultured cells, and luciferase activity was assayed. These data compare the degree of expression of luciferase with differences in the location of DNADNA methylation [Irvine et al. (2002). Mol. and Cell. Biol. 22:6689−−6696]. What general conclusions can be drawn from these data?
Methylation outside of the transcriptional unit results in a small reduction in luciferase expression, whereas methylation within the unit causes a drastic reduction in expression. The effect of methylation of the transcriptional unit may be enhanced by methylation outside the unit.
Which of the following mutations would result in cancer?
Mutations that prevents the retinoblastoma protein (pRB) from binding to E2F transcription factor. Mutation that produces cyclin dependent kinases that are permanently active, regardless of cyclin levels.
If an individual with the minimum height specified by these genes marries an individual of intermediate or moderate height, will any of their children be taller than the tall parent? Why or why not?
No, there is no way of having more than four uppercase alleles in the offspring.
Indicate possible sets of genotypes for the original P1P1 parents and the F1F1 plants that could account for these results.
P1: aaBBCCdd×AAbbccDD aaBBccDD×AAbbCCdd AABBccdd×aabbCCDD F1: AaBbCcDd×AaBbCcDd
In a herd of dairy cows the narrow-sense heritability for milk protein content is 0.76, and for milk butterfat it is 0.82. The correlation coefficient between milk protein content and butterfat is 0.91. If the farmer selects for cows producing more butterfat in their milk, what will be the most likely effect on milk protein content in the next generation?
Protein content will increase
Calculate the overall realized heritability.
The average h2 = 1.139
Describe the roles of mutation, migration, and selection in bringing about speciation.
The original sources of variation coming from mutation. Migration can cause gene frequencies to change in a population if the immigrants have different gene frequencies compared to the host population. Then by selection genetic mutation that enhance reproduction become and remain more common in successive generations of a population. Under certain conditions, population that at one time could interbreed may lose that capability, thus segregating their adaptations into particular niches. If selection continues, it may result in the appearance of new species.
The percentage of heterozygous carriers can be calculated as 2×(p0.5)×(1−p0.5)×1002×(p0.5)×(1−p0.5)×100. What must be assumed in order to use this formula?
The population must be in Hardy-Weinberg equilibrium.
You are investigating a new bacterial operon that contains five regions (A, B, C, D and E) involved in coordinated regulation of transcription. One is the gene for a regulatory protein, one is the binding site for the regulatory protein, two produce structural enzymes (enzyme 1 and enzyme 2), and one is the promoter for the two enzyme genes.The table below shows data collected for five different bacterial strains. A "+" indicates that the structural enzyme is produced and a "-" indicates that the enzyme is not produced. The "signal" is a molecule that either represses or induces the operon. Use the information to determine how the operon is regulated, and the identity of each region. In the wild-type operon (A+ B+ C+ D+ E+), how does the presence of the signal affect expression of the structural enzymes ?
The presence of the signal stops production of the structural enzymes. Without the signal, the structural enzymes are produced.
A number of comparisons of nucleotide sequences among hominids and rodents indicate that inbreeding may have occurred more often in hominid than in rodent ancestry. Bakewell et al. (2007. Proc. Nat. Acad. Sci. [USA] 104: 7489−−7494) suggest that an ancient population bottleneck that left approximately 10,000 humans might have caused early humans to have a greater chance of genetic disease. Why would a population bottleneck influence the frequency of genetic disease?
When a population bottleneck occurs and the number of effective breeders is reduced in a population, two phenomena usually follow. First, because the population is small, wide fluctuations in genotypic frequencies occur, thereby revealing deleterious alleles by chance. Second, inbreeding often occurs in small populations, thereby increasing the chance for homozygosity. With increased homozygosity comes a/an increased likelihood that recessive alleles will be expressed. Since many disease-producing genes are recessive, a/an increase in genetic diseases is a likely aftermath to a population bottleneck.
Are these overlaps explained by different modifications?
Yes, except the third methylation of lysine 79, which can result in either activation or repression.
Should researchers be looking for mutant alleles of genes that control formation and differentiation of the neural tube?
Yes, it may increase our understanding of not only spina bifida but also many other conditions that result from neural tube defects.
Are there any overlapping modifications?
Yes, modification of lysine residues 9, 27, and 79 can cause either activation or repression of expression.
Height in humans depends on the additive action of genes. Assume that this trait is controlled by the four loci R, S, T, and U and that environmental effects are negligible. Instead of additive versus nonadditive alleles, assume that additive and partially additive alleles exist. Additive alleles contribute two units, and partially additive alleles contribute one unit to height. Can two individuals of moderate height produce offspring that are much taller or shorter than either parent? If so, how?
Yes, the example of the crossing is: rrSsTtuu ×× RrSsTtUu
Now that you have identified the number of genotypes predicted for a gene that has three alleles, can you calculate phenotype frequencies when just allele frequencies are known? For the ABO blood group in humans, there are three alleles (IA , IB , and i), six possible genotypes (IAIA , IBIB , IAIB , IAi, IBi, and ii), and four possible phenotypes (A, B, AB, and O). Recall that IA is dominant to i, IB is dominant to i, and IA and IB are codominant. In a given population, the allele frequencies are as follows: IA = 0.15IB = 0.25i = 0.60
a = 0.2025 b= 0.3625 c= 0.075 o= 0.36
Assuming that the realized heritability in phlox is relatively high, what factors might account for such a high response?
a few number of loci involved great genetic variability
In this problem, you will explore how gene flow (migration) affects allele frequencies in island populations. A mainland and an island are both inhabited by brown and white mice. Allele A1 codes for brown coat color; allele A2 codes for white coat color. A1 is dominant to A2 .Mice do not swim between the mainland and the island, and there are no shipping routes to the island, thus mice cannot hitch a ride on ships. During storms, however, some mice get "rafted" to the island on logs.The island contains 240 mice. The frequency of the A1 allele is 0.85; the frequency of the A2 allele is 0.15.On the mainland, the frequency of the A1 allele is 0.55; the frequency of the A2 allele is 0.45.During a severe storm, 60 mice from the mainland are rafted to the island. What are the new frequencies of A1 and A2 on the island after the storm?
frequency of A1 , frequency of A2 0.790,0.210
Consider a population of frogs. Allele B1 codes for spotted frogs; allele B2 codes for non-spotted frogs. B1 is dominant to B2 . The allele frequencies on the mainland are B1 = 0.75 and B2 = 0.25. The allele frequencies on the island are B1 = 0.55 and B2 = 0.45. There are 170 frogs on the island. During a storm, 30 non-spotted frogs come to the island by floating on debris. What are the new frequencies of B1 and B2 on the island?
frequency of B1 , frequency of B2 0.470,0.530
Calculate the realized heritability for 1998 year.
h2 = 0.979
Calculate the realized heritability for 1999 year.
h2 = 1.906
Which of the following mutations would result in higher-than-normal expression of the trp genes in presence of tryptophan?
mutations in Region 3 that prevent 3-4 stem loop formation, mutations in Region 1 that prevent 1-2 stem loop formation
The trp and lac operons are regulated in various ways. How do bacteria regulate transcription of these operons?
operon is not transcribed- 1. lac operon: lactose absent 2.trp operon: tryptophan present operon is transcibed, but notsped up through positive control 1.trp operon: tryptophan absent 2.lac operon: lactose present,glucose present operon is transcribed quickly through positive control 1.lac operon: lactose present,glucose absent
In a population of 60 desert bighorn sheep, a mutant recessive allele cc when homozygous causes curled coats in both males and females. The normal dominant allele CC produces straight coats. A biologist studying these sheep counts four with curled coats. She also takes blood samples from the population for DNA analysis, which reveals that 19 of the sheep are heterozygous carriers of the cc allele. To increase genetic diversity in this bighorn sheep population, thirty sheep are introduced from a population where the cc allele is absent. Assuming that random mating occurs between the original and the introduced sheep, and that the cc allele is selectively neutral, what will be the frequency of cc in the next generation?Recall that the equation for determining the impact of immigration on the gene pool of an existing population is estimated by the following equation:pi′=(1−m)pi+mpmpi′=(1−m)pi+mpmwhere m represents migrants.
pi = 0.89
An inbred strain of plants has a mean height of 24 cmcm. A second strain of the same species from a different geographical region also has a mean height of 24 cmcm. When plants from the two strains are crossed together, the F1F1 plants are the same height as the parent plants. However, the F2F2 generation shows a wide range of heights; the majority are like the P1P1 and F1F1 plants, but approximately 4 of 1000 are only 12 cmcm high, and about 4 of 1000 are 36 cmcm high. What mode of inheritance is occurring here?
polygenic
When does allele frequency equilibrium occur?
when ΔΔ p I = 0, when m(p C - p I) = 0, when m = 0
Present an overview of the manner in which chromatin can be remodeled. Describe the manner in which these remodeling processes influence transcription.
1. DNA methylation has been linked to a decrease(s) in gene expression. It most often occurs at position 5 of cytosine. 2. Histone acetylation decrease(s) the positive charge on histones, resulting in a reduced affinity of the histone for DNA. The process is catalyzed by HATs and results in forming "open" chromatin. 3. Chromatin remodeling leaves DNA open to associate with transcription factors and RNA polymerase by repositioning or removing nucleosomes using large, energy dependent multi-subunit complexes. 4. One way in which nucleosomes can be modified is by changing their composition. Histone variants, like H2A.Z affect nucleosome mobility and positioning on DNA.
As a genetic counselor, you are asked to assess the risk for a couple with a family history of familial adenomatous polyposis (FAP) who are thinking about having children. Neither the husband nor the wife has colorectal cancer, but the husband has a sister with FAP. What is the probability that this couple will have a child with FAP?
1/4
How many gene pairs are involved
4
How does the Ras protein transmit a signal from outside the cell into the cytoplasm?
Activated Ras proteins transduce a signal, which activates the transcription of genes that start cell division.
What is the evidence that epigenetic changes are involved in cancer?
All cancers tested thus far have exhibited a pattern of global hypomethylation, although specific regions are hypermethylated. The genes that are silenced include tumor suppressors.
Which of the following statements regarding heritability of cancer is true?
At least two mutational events are required for a cell to become cancerous.
Why?
Because the banding pattern in the patient's sample matches that of his mother and includes both maternal STR alleles.
The period of greatest susceptibility to diseases resulting from epigenetic changes is ______________________.
Before birth
Choose the correct explanation of the apparent paradox that both hypermethylation and hypomethylation of DNA are often found in the same cancer cell.
Complex regulatory systems are involved. For example, hypomethylation of oncogenes and hypermethylation of tumor-suppressor genes are expected in cancer cells.
From the information on a mode of inheritance referred to as quantitative genetics, as well as many of the statistical parameters utilized to study quantitative traits, what answers would you propose to the following fundamental questions. How do we know that threshold traits are actually polygenic even though they may have as few as two discrete phenotypic classes?
Environmental factors have a considerable impact on polygenic system, and therefore, on expression of threshold traits. Polygenic conditions involving thresholds occur if a range of expression is possible, but the trait is either expressed or not expressed.
How does an environmental factor like stress generate a response that is transmitted from generation to generation?
Environmental factors induce epigenetic changes in genes that result in certain types of behavior. Since it has been shown to be heritable, it is believed that the influence of environmental factors can also be passed on to subsequent generations.
Epigenetics is a relatively new area of genetics with a focus on phenomena that affect gene expression but do not affect DNA sequence. Epigenetic effects are quasi-stable and may be passed to progeny somatic or germ-line cells. What are known causes of epigenetic effects, and how do they relate to cancer?
Epigenetic effects can be caused by DNA methylation and/or histone acetylation and/or phosphorylation. As such, they can silence or activate chromosomes or certain chromosomal regions and be responsible for parental imprinting or influencing gene activity in heterochromatin. Patterns of nucleotide demethylation and hypermethylation are often different when cancer cells are compared to normal cells
What findings led geneticists to postulate the multiple-factor hypothesis that invoked the idea of additive alleles to explain inheritance patterns?
Experimental results involving pigmentation in wheat showed that a number of additive alleles acting in Mendelian fashion could explain continuous variation.
Under what circumstances do cells undergo apoptosis?
If DNA or chromosomal damage is so severe that its repair is impossible, the cell may initiate programmed cell death.
Identify how different types of histone modification impact transcription.
Increases gene expression: histone acetyltransferases (HATs), acetylation, decreased attraction between DNADNA and histones
Given that TP53TP53 is a recessive gene and is not located on the X chromosome, why would people who inherit just one mutant copy of a recessive tumor-suppressor gene be at higher risk of developing cancer than those without the recessive gene?
Individuals with two copies of the gene would need to experience separate mutations in both copies to develop cancer, whereas individuals with one functional copy would only need a single mutation.
What happens in cases where the rasras gene is mutated?
It continually signals cell division.
What is apoptosis?
It is a programmed cell death, a genetically controlled process that leads to the death of a cell.
What is the role of the p53 protein in the cell cycle in normal cells?
It temporarily arrests the cell cycle in G1 before entering S.
What are the similarities and differences in these two types of ncRNAncRNAs?
Long ncRNAs prevent transcription by chromatin remodeling, forming chromosomal loops etc. interact with chromatin regulators Short ncRNAs: interact with RISC silence the targeted messages by cleaving them or by blocking translation Both operate by forming complexes act as guides for RNARNA-protein complexes
Does this exclude genetic mutations as a cause of this condition?
No, this does not exclude genetic causes. While taking folic acid prevents many cases of spina bifida, it does not prevent all of them.
What is the relationship between proto-oncogenes and oncogenes?
Oncogenes are mutant forms of proto-oncogenes.
How does this property of lysine allow it to interact with DNA?
Positively charged lysine side chains can form ionic bonds with negatively charged phosphate groups in the DNADNA backbone.
Radiotherapy (treatment with ionizing radiation) is one of the most effective current cancer treatments. It works by damaging DNA and other cellular components. In which ways could radiotherapy control or cure cancer?
Radiotherapy is often administered externally or internally to kill cancer cells by damaging their DNA.
Parts of an operon that control the expression of both enzyme genes are called "regulatory regions." The genes that produce the enzymes are called "structural genes." Using the data in the table, determine which regions are regulatory regions and which are structural genes.
Regulatory Region(s): Region B Region D Region A Structural Gene(s): Region E Region C
Identify how different types of histone modification impact transcription.
Repositions histones: chromatin remodeling complexes
Which of the following might explain why the level of cyclin B in a cell is not increasing heading into mitosis?
Spindle fibers are not properly formed. The attachment of spindle fibers to the kinetochores is inadequate.
Which of the following statements describes metastasis?
The ability to form secondary tumors at another site
How can the role of epigenetics in cancer be reconciled with the idea that cancer is caused by the accumulation of genetic mutations in tumor-suppressor genes and proto-oncogenes?
The accumulation of genetic mutations in tumor-suppressor genes and proto-oncogenes changes their expression or causes the production of nonfunctional products. Epigenetic modifications can also change gene expression and, therefore, can cause cancer.
Identical twins each carry the same genome, but over time, can develop different phenotypes. How can you explain this?
The environmental exposures lead to differences in their epigenome and, therefore, differences in gene expression.
Several mechanisms are involved in epigenetic modifications to the genome that regulate gene expression, and epigenetic control of gene expression is important in development, cancer, and modulating the genomic response to environmental factors. Using your knowledge, answer the following questions. How do we know how methylation of promoters silences gene expression?
The methyl groups sit in the major groove and prevent binding of transcription factors and other proteins needed to form an initiation complex.
Now that you have determined which are the regulatory regions, identify how this operon is regulated.
The operon is repressible (the signal represses/stops expression of the enzymes). The operon is under negative control (the regulatory protein is a repressor).
If the UGG codons in Region 1 of trpL were changed to AGG codons, what effect would this have on expression of the trp operon?
The operon would be regulated by arginine levels; high levels of arginine would attenuate expression of trp genes.
How can you reconcile these differences?
The result of the third methylation of lysine 79 may depend on the interaction of this modified amino acid with other modified residues.
Although the mutations listed in the table are clearly deleterious and cause breast cancer in women at very young ages, each of the kindred groups had at least one woman who carried the mutation but lived until age 80 without developing cancer. Choose mechanisms (or variables) that could underlie variation in the expression of a mutant phenotype, and propose an explanation for the incomplete penetrance of this mutation.
Their immune systems are more efficient at destroying precancerous cells. They may have mutations in breast signal transduction genes so that suppression of cell division occurs in the absence of BRCA1BRCA1. They may carry genes (perhaps mutant) whose products provide functions that are the same as that of the BRCA1BRCA1 gene product.
What are the functions of lncRNAlncRNAs in epigenetic regulation?
They can bind transcription factors, effectively sequestering them from target genes (the decoy strategy) They can facilitate the assembly of proteins that will then act to regulate gene expression (the adapter strategy) They can recruit chromatin remodeling complexes to the DNA in an allele-specific fashion (the guide strategy) They can induce chromosome looping to allow for gene regulation (the enhancer strategy)
Which of the following is NOT a characteristic of retroviruses?
They cause lysis of the host cell upon infection.
Our environment contains both natural and human-made carcinogens.
True
How do we assess environmental factors to determine if they impact the phenotype of a quantitatively inherited trait?
by heritability estimation
All of the following are characteristics of genomic instability in cancer cells EXCEPT __________.
controlled cell division
How do we know that monozygotic twins are not identical genotypically as adults?
differences in inherited disease states
What are the major mechanisms of epigenetic genome modification?
histone modification involving acetyl, methyl, and phosphate groups chromatin remodeling binding of RISC to target mRNA molecules long noncoding RNAs binding to chromatin-modifying enzymes DNA methylation
What are the major mechanisms of epigenetic genome modification?
histone modification involving acetyl, methyl, and phosphate groups long noncoding RNAs binding to chromatin-modifying enzymes chromatin remodeling binding of RISC to target mRNA molecules DNA methylation
When are the major regulatory points in the cell cycle?
late G1 phase (G1/S checkpoint) late G2 phase (G2/M checkpoint) M phase (M checkpoint)
Predict the outcome of crossing a true-breeding medium-red plant and a white plant. What color(s) would you expect in the F1 offspring?
light-red
Which H3 modifications are associated with repression?
lysine 9 lysine 27 lysine 79 serine 10
Given only the data in the table, for which part(s) of the operon is it possible to identify the exact region it corresponds to?
the gene for enzyme 2, the promoter region, the gene for enzyme 1
Examine the types of mutations that are listed in the table, and determine if the BRCA1BRCA1 gene is likely to be a tumor-suppressor gene or an oncogene.
tumor-suppressor gene
Predict the outcome of crossing two of the F1 plants from Part D. What color(s) would you expect in the offspring?
white, light-red, medium-red
A dark-red strain and a white strain of wheat are crossed and produce an intermediate, medium-red F1.When the F1 plants are interbred, an F2 generation is produced in this ratio:1 dark-red: 4 medium-dark-red: 6 medium-red: 4 light-red: 1 whiteFurther crosses reveal that the dark-red and white F2 plants are true breeding. Based on the ratios in the F2 population, how many genes are involved in the production of color?
2 genes
How much does each gene contribute to the plant height?
3 cm
Note the coding effect of the mutation found in kindred group 2082. This results from a single base-pair substitution. Choose the normal double-stranded DNADNA sequence for this codon (with the 5′5′ and 3′3′ ends labeled) and the sequence after the mutation that resulted from an uncorrected mismatch event during DNADNA replication.
5′-CAG(A)-3′5′-CAG(A)-3′ coding3′-GTC(T)-5′3′-GTC(T)-5′ templatechanged into5′-TAG(A)-3′5′-TAG(A)-3′ coding3′-ATC(T)-5′3′-ATC(T)-5′ template
Are there any tests that you could recommend to help in this assessment?
A colonoscopy can be used on the husband to detect presence of polyps.
Use the information from Part B to determine which genotype(s) in the F2 generation would produce an ear length of 13.5 cm.
AaBB, AABb
How does acetylation of lysine affect its interaction with DNADNA, and how is this related to the activation of gene expression?
Acetylation of the lysine side chain removes its positive charge preventing the ionic interaction with phosphate and decreasing the interaction between the histone and the DNA. As a result, genes are more accessible and able to be expressed.
What are the differences and similarities among the three classes of monoallelic gene expression?
Genomic imprinting: is dependent on the parent that contributed the allele is found less than in 1 percent of the genome and usually found in clusters Inactivation of the X chromosome: is mediated by the coordinated expression of two antisense long noncoding RNAs occurs only in females the entire chromosome is inactivated Inactivation of autosomal genes: is found in only some cells, resulting in a mosaic pattern of expression in an individual is found in 15-20 percent of genes All classes of monoallelic gene expression: once a gene has been inactivated, it remains inactive in that individual involves the inactivation of one of the two inherited alleles
Under what conditions do the different stem-loop structures occur, and what effect do they have on transcription of the trp genes?
High tryptophan Region 1 - Region 2 stem loop Region 3 - Region 4 stem loop trp genes not transcribed Low tryptophan Region 2 - Region 3 stem loop trp genes transcribed
Identify how different types of histone modification impact transcription.
Decreases gene expression: histone deacetylases (HDACs), deacetylation, increased attraction between DNADNA and histones
Based on your interpretation of the data, what is the cause of PWS in this case?
loss of the paternal copy of chromosome 15 uniparental disomy
From the data in the table, which histone H3 modifications are associated with gene activation?
lysine 36 lysine 14 lysine 27 lysine 79 lysine 9 lysine 4
Many transcriptional activators are proteins with a DNADNA-binding domain (DBD) and an activation domain (AD). Explain how each domain contributes to transcriptional initiation.
1. The DBD allows the activator to bind to enhancer elements, to correctly position the protein to interact with the correct gene sequence. 2. The AD allows the activator to interact with other proteins, such as coactivators, important to the formation of the preinitiation complex.
Find the mean height of a population of palm trees with the following information: Trees with heights of 140 feet are bred, and the average height of the progeny is 128 feet. The selection response, R, is 70, and the selection differential, S, is 100
100
Suppose that ear length in rabbits is controlled by two additive genes, each of which has two alleles. A true-breeding female (aabb) with 6-cm ears is mated to a true-breeding male (AABB) with 16-cm ears.Predict the ear length of the F1 generation. (Note: Assume that each allele contributes equally to ear length, and the genes segregate independently. Also assume that environmental influence on ear length is negligible.)
11 cm
The significance of checkpoints can be demonstrated by considering what happens when they are impaired. What would occur if there was a gain-of-function mutation in the promoter for the cyclin E gene such that cyclin E protein was always made at high levels even under conditions in which cyclin E would not normally be made?
Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.
Does it seem likely that imprinting disorders could be treated by controlling the maternal environment in some way, perhaps by dietary changes?
Evidence suggests that maternal nutrition does not largely influence global methylation patterns, particularly in nutrient-replete populations; however, an important impact on gene-specific methylation is observed.
Expression of the trp operon is regulated by the level of free tryptophan in the cell both through repressor action and attenuation. This chart shows the percent expression of the trp operon in the presence and absence of tryptophan for wild-type (trpR+ or trp+) and repressor mutants (trpR-).Use this information to determine the levels of expression for the following genotypes and conditions.
Highest to Lowest: trp+ / tryptophan absent trpL- / tryptophan present trpR- / tryptophan present trp+ / tryptophan present trpP- / tryptophan absent
Even though cancer genomes are overall hypomethylated, explain why some genes are hypermethylated in cancer cells.
Hypermethylation of promoters will lead to reduced expression of the correspondent genes. Some genes such as cell cycle stop genes and DNA repair genes need to be expressed. When they are not expressed, this can lead to cancer
How do normal cells protect themselves from accumulating mutations in genes that could lead to cancer? How do cancer cells differ from normal cells in these processes?
Normal cells are often capable of withstanding mutational assault because they have checkpoints and DNA repair mechanisms in place. When such mechanisms fail, cancer may be a result. Through mutation, such protective mechanisms are compromised in cancer cells, and as a result, they show higher than normal rates of mutation, chromosomal abnormalities, and genomic instability
You are studying a bacterium that utilizes a sugar called athelose. This sugar can be used as an energy source when necessary. Metabolism of athelose is controlled by the ath operon. The genes of the ath operon code for the enzymes necessary to use athelose as an energy source. You have found the following: The genes of the ath operon are expressed only when the concentration of athelose in the bacterium is high. When glucose is absent, the bacterium needs to metabolize athelose as an energy source as much as possible. The same catabolite activator protein (CAP) involved with the lac operon interacts with the ath operon. Based on this information, how is the ath operon most likely controlled?
Positive control: inactive activator + small molecule = active activator Negative Control: active repressor + small molecule = inactive repressor
Because the degree of DNADNA methylation appears to be a relatively reliable genetic marker for some forms of cancer, researchers have explored the possibility of altering DNADNA methylation as a form of cancer therapy. Initial studies indicate that while hypomethylation suppresses the formation of some tumors, other tumors thrive. Why would one expect different cancers to respond differently to either hypomethylation or hypermethylation therapies?
Since there are multiple routes that lead to cancer, one would expect complex regulatory systems to be involved.
Mutations in region B and region D give the same results - when either of these regions is mutated, the operon is expressed under all conditions (called constitutive expression).You hypothesize that region B is the operator region and that Region D is the repressor protein.To test your hypothesis, you create two partial diploid lines by introducing a F' plasmid with a wild-type lac operon:Strain 1: F' A+ B+ C+ D+ E+ / A+ B- C+ D+ E+Strain 2: F' A+ B+ C+ D+ E+ / A+ B+ C+ D- E+ What experimental results would be predicted by your hypothesis?
Strain 1 Left to Right: +,+,+,+ Strain 2 Left to Right: +, -, +, -
What are the three classes of monoallelic gene expression?
random inactivation of autosomal genes random inactivation of the X chromosome genomic imprinting
Choose the two types of ncRNAncRNAs involved in epigenetic control of gene expression.
short ncRNAncRNAs and long ncRNAncRNAs
The mRNA that is transcribed from the trpL region can alternately fold to form three possible stem-loop structures. Different stem-loop structures have varying effects on transcription. Which stem-loop structure is responsible for terminating transcription?
the 3-4 stem loop
Measurements of serum cholesterol are shown for monozygotic twins that were separated at birth: one was reared in Glasgow and the other reared in London. From these studies like these, the broad sense heritability can be calculate using the formula: H2 = COVx'x"/Vx In this fomula, COVx'x" is the covariance where the those from Glasgow make up the first array and those from London are the second array. Calculate the covariance in the study. These four sets of twins are a sample of the overall population.
280
Which conditions must be met before the cell can begin DNA replication?
Cyclin D1 levels must be high The transcription factor, E2F, must be bound to DNA. The transcription factor, E2F, must be bound to DNA.
Why does radiotherapy often have significant side effects?
In addition to killing cancer cells, radiotherapy can also kill normal cells too, causing side effects.
Erythritol, a natural sugar abundant in fruits and fermenting foods, is about 65 percent as sweet as table sugar and has about 95 percent fewer calories. It is "tooth friendly" and generally devoid of negative side effects as a human consumable product. Pathogenic Brucella strains that catabolize erythritol contain four closely spaced genes, all involved in erythritol metabolism. One of the four genes (eryDeryD) encodes a product that represses the expression of the other three genes. Erythritol catabolism is stimulated by erythritol. Present a simple regulatory model to account for the regulation of erythritol catabolism in Brucella.
The product of the eryD gene is probably involved in repressing a promoter site. The system is induced when erythritol interrupts the action of the eryD repressor.
What features of eukaryotes provide additional opportunities for the regulation of gene expression compared to bacteria?
separated processes of transcription and translation opportunity for posttranscriptional and pre-translational regulation presence of chromatin
Suppose that tail length in horses is controlled by a number of additive genes, each of which has 2 alleles.When a true-breeding female with a 12-inch tail is mated to a true-breeding male with a 36-inch tail, all F1 offspring have the same tail length. When the F1 offspring are mated to each other, 7 different tail lengths are seen in the F2 offspring.How many genes contribute to tail length? (Note: Assume that the genes segregate independently and that environmental influence on tail length is negligible.)
3 genes
Measurements of serum cholesterol are shown for monozygotic twins that were separated at birth: one was reared in Glasgow and the other reared in London.From these studies like these, the broad sense heritability can be calculate using the formula: H2 = COVx'x"/Vx In this fomula Vx is the variability of all individuals in the study. Calculate the variability for all 8 individuals in the study. These eight individuals are a sample of the overall population. SampleGlasgowLondon 222 228 172 196 204 220 200 217
334
Prader-Willi syndrome (PWS) is a genetic disorder with a clinical profile of obesity, intellectual disability, and short stature. It can be caused in several ways. Most common is a deletion on the paternal copy of chromosome 15, but it can also be caused by an epigenetic imprinting disorder, and uniparental disomy, an event in which the affected child receives two copies of the maternal chromosome 15. A child with PWS comes to your clinic for a diagnosis of the molecular basis for this condition. The gel below shows the results of testing with short tandem repeats (STRs) from the region of chromosome 15 associated with the disorder. Is this case caused by a deletion in the paternal copy of chromosome 15?
No, the disorder was not caused simply by a deletion in the paternal copy of chromosome 15. If this were the case, the gel results for the patient would show a single, maternal STR allele and no paternal STR allele.
